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Leona [35]
2 years ago
13

Which model shows 1/2 times 1/3

Mathematics
2 answers:
Bess [88]2 years ago
8 0

Hai there!

First, let's use the KCF technique (my old math teacher taught me this!)

Keep the first fraction the same, in this case, 1/2

Change the sign from division to multiplication

Flip the second fraction, or 3/8, to get it's reciprocal, or 8/3.

The new equation should look like this:

1/2 * 8/3

Now multiply straight across like you normally would, like so:

1 * 8 = 8

2 * 3 = 6

arlik [135]2 years ago
7 0

Answer:

Where's the model?

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Pleaseee helppp with thissss asapppp
Natasha2012 [34]

As we know that the standard equation of circle is {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}} , where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>

Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}

{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}

{:\implies \quad \sf 9+16=r^{2}}

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>

Now , putting values in our standard equation ;

{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}

<em>This is the required equation of </em><em>Circle</em>

Refer to the attachment as well !

8 0
2 years ago
Solve the system by substitution. y = 8x + 32 Y= -8x​
leonid [27]
The answer is (x,y) = (-2, 16)

do you need an explanation as well?
6 0
3 years ago
-<br><img src="https://tex.z-dn.net/?f=%20-%203%5Cpi%20%2B%20w%20%3D%202%5Cpi" id="TexFormula1" title=" - 3\pi + w = 2\pi" alt="
pantera1 [17]

Isolate the w. Note the equal sign. What you do to one side, you do to the other.

Add 3π to both sides

- 3π (+3π) + w = 2π (+3π)

w = 2π + (3π)

w = 5π

w = 5π is your answer

hope this helps

4 0
3 years ago
Which of the following is a factor of 2x^4 22x^3 60x^2? 2x^3 x^4 x 4 x 5
zavuch27 [327]
The factor theorem says that for a function f(x), if f(a) = 0, then (x - a) is a factor of f(x).
For f(x) = 2x^4 + 22x^3 + 60x^2
f(-5) = 2(-5)^4 + 22(-5)^3 + 60(-5)^2 = 2(625) + 22(-125) + 60(25) = 1250 - 2750 + 1500 = 0
Therefore, x - (-5) = x + 5 is a factor of f(x).
8 0
4 years ago
Please answer the question to the best of your ability.
nydimaria [60]

Answer:

the answer is c.

Step-by-step explanation:

apply the rule of distributive property to check each of the options provided

7 0
3 years ago
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